Balancing and Yield GCHEM question

[pandalove89]

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(MnO4)^- + I^- + H^+ --> Mn^2+ + H2O + I2

If 150 mg of KI is reacted and 108.5 mg of I2 is produced what is the percent yield for the reaction

So I balanced it out to 2MnO4^- + 10I^- + 16H --> 2Mn^2+ + 5I2 + 8H2O

but I have no idea where to go from here

 

[pandalove89]

bump

 

[PooyaH]

This is just a %yield type of problem.

150mg KI (1mol/167g KI)(5mol I2/10mol KI)(254g/1mol I2) = 114g I2 (this is your theoretical yield)

108.5g/114g = .95g I2 so you have a 95% yield.

 

[Hope30]

I will take a stab at this.
Assuming your balanced equation is right,

150mg * (1g/1000mg) * (1mol KI/ 166g KI) * (1mol I-/ 1mol KI) * (5mol I2/ 10mol I-) * (254g I2/ 1mol I2) = .115 g = 115mg I2

108.5mg is what you obtain experimentally and 115mg is your theoretical amount, so 108.5/115 = 0.94 or 94%

94% is your percent yield

 

[Hope30]

lol take me longer than I thought to type this. Did not see that.